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zalisa [80]
4 years ago
9

The Greek alphabet has 24 distinct lowercase letters. How many bits are needed to be able to encode any single lowercase Greek l

etter
Engineering
1 answer:
Alexus [3.1K]4 years ago
4 0

Answer:

You need 5 bits to encode al the 24 greek distinct lowercase letters.

Explanation:

o encode 24 greek distinct lowercase letters, you should be able to write in binary language 24 different combinations.

With 1 bit you have only 2 possible combinations. Not enough.

With 2 bit you have only 4 possible combinations. Not enough.

With 3 and 4 bit you have only 8 and 16 possible combinations. Not enough.

With 5 bit you have 32 possible combinations. This is enough to encode all the 24 Greek letters.

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An ice hockey player is skating on an ice rink. The rink has a coefficient of kinetic friction of roughly 0.1. If the normal for
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3 years ago
Please help me with this question​
kap26 [50]

Answer:

The current through each lamp is 0.273 Amperes

Power dissipated in each lamp is 0.082W

Explanation:

Battery v = 1.5 V

Each lamp has resistance, r = 1.1 Ohms

The 5 lamps in series will therefore have total resistance, R = 5 * 1.1 = 5.5 Ohms

The current through each lamp, I = v/R = 1.5 / 5.5 = 0.273 Amperes

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3 0
3 years ago
The cantilevered W530 x 150 beam shown is subjected to a 9.8-kN force F applied by means of a welded plate at A. Determine the e
snow_lady [41]

The <em>equivalent force-couple system</em> at O is the force and couple experienced when at point O due to the applied force at point A

The <em>equivalent force couple</em> system at O due to force <em>F</em> are;

Force, F =  (<u>8.65·i - 4.6·j</u>) KN

Couple, M₀ ≈ <u>40.9 </u>k kN·m

The reason the above values are correct is as follows:

The known values for the <em>cantilever</em> are;

The <em>height </em>of the beam = 0.65 m

The <em>magnitude of </em>the applied <em>force</em>, F = 9.8 kN

The <em>length </em>of the beam = 4.9 m

The <em>angle </em>away from the vertical the force is applied = 26°

The required parameter:

The <em>equivalent force-couple system</em> at the centroid of the beam cross-section of the cantilever

Solution:

The <em>equivalent force-couple system</em> is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;

The <em>equivalent force</em> \overset \longrightarrow F = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j

Which gives;

The <em>equivalent force</em> \overset \longrightarrow F ≈ (<u>8.65·i - 4.6·j</u>) KN

The <em>couple </em><em>acting </em>at point O due to the force <em>F</em> is given as follows;

The <em>clockwise moment</em> = <em>9.8 kN × cos(28°) × 4.9</em>

The <em>anticlockwise moment</em> = <em>9.8 kN × sin(28°) 0.65/2 </em>

The sum of the moments = Anticlockwise moment - Clockwise moments

∴ The <em>sum </em>of the moments, ∑M, gives the moment acting at point O as follows;

M₀ = <em>9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9</em>  ≈ 40.9 kN·m

The couple acting at O, due to F,  M₀ ≈ <u>40.9 kN·m</u>

The equivalent force couple system acting at point O due the force, F, is as follows

F =  (8.65·i - 4.6·j) KN

M₀ ≈ <u>40.9 </u>k kN·m

Learn more about equivalent force systems here:

brainly.com/question/12209585

4 0
3 years ago
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sergey [27]

Answer:

See attached picture.

Explanation:

See attached picture for explanation.

4 0
3 years ago
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